Find this $a,b,c$ such that $\sqrt{9-8\sin 50^{\circ}}=a+b\sin c^{\circ}$

We have the following (working in degrees):

$$\cos 20 - \cos 80 = \cos(50-30) - \cos(50+30) = 2 \sin 50 \sin 30 = \sin 50$$

Thus we have that

$$1 - 2\sin^2 10 - \sin 10 = \sin 50$$

(using $\cos 20 = 1 - 2 \sin^2 10$ and $\cos 80 = \sin (90 - 80) = \sin 10$)

And so

$$9 - 8 \sin 50 = 9 - 8(1 - 2\sin^2 10 - \sin 10) = 1 + 8\sin 10 + 16\sin^2 10 = (1 + 4 \sin 10)^2$$

Thus $a=1, b=4, c=10$.